Analysis: (1) It is found that ∠ bam+∠ ABN = 270; ∠MAB is divided equally by Associated Press, and∠ ABN is divided equally by BP. ∴∠ BAP+∠ ABP =

(∠ bam +∠ ABN) = 135, based on the theorem ∠ p =180-135 = 45;

(2) Same thinking as the question (1);

(3) Using the properties of external angles, we can get ∠ BAM+∠ ABN = ∠ MON+∠ BAO = (∠ MON+∠ ABO+∠ BAO) +∞. Divide ∠MAB by AP and ∠ ABN by BP. Find ∠BAP+∠ABP, and find the degree ∠ BAP+∠ ABP+∠ P = 180 by using the theorem of triangle interior angle sum.

Solution: (1)∵∠BAM is the external angle of △AOB.

∴∠BAM=∠AOB+∠ABO

∫∠ABN is the outer corner of △AOB.

∴∠ABN=∠AOB+∠BAO

∴∠bam+∠abn=∠aob+∠abo+∠aob+∠bao=(∠aob+∠abo+∠bao)+∠aob= 180+90 = 270

∫ Associated Press stock ∝∠MAB, BP stock ∝∠ABN.

∴∠BAP= 1/2

∠BAM，∠ABP=

1/2∠ABN

∴∠BAP+∠ABP=

1/2 (bam+ABN) =135

At △ABP

∠BAP+∠ABP+∠P= 180

∴∠p= 180- 135 = 45；

(2) 𕚸 BAM is the outer corner of △AOB.

∴∠BAM=∠AOB+∠ABO

∫∠ABN is the outer corner of △AOB.

∴∠ABN=∠AOB+∠BAO

∴∠bam+∠abn=∠aob+∠abo+∠aob+∠bao=(∠aob+∠abo+∠bao)+∠aob= 180+80 = 260

∫ Associated Press stock ∝∠MAB, BP stock ∝∠ABN.

∴∠BAP= 1/2

∠BAM，∠ABP=

1/2∠ABN

∴∠BAP+∠ABP=

1/2(∠ bam +∠ABN)= 130

At △ABP

∠BAP+∠ABP+∠P= 180

∴∠p= 180- 130 = 50；

(3)∠MON+2∠P= 180

∫∠BAM is the outer corner of △AOB.

∴∠BAM=∠MON+∠ABO

∫∠ABN is the outer corner of △AOB.

∴∠ABN=∠MON+∠BAO

∴∠ BAM+∠ ABN = ∠ MON+∠ MON+∠ BAO = (∠ MON+∠ ABO+∠ BAO)+∠ MON =

∫ Associated Press stock ∝∠MAB, BP stock ∝∠ABN.

∴∠BAP=

1/2∠BAM，∠ABP=

1/2∠ABN

∴∠BAP+∠ABP=

1/2(∠ bam +∠ABN)= 0

1/2( 180 +∠MON)

At △ABP

∠BAP+∠ABP+∠P= 180

1/2( 180+∠MON)+∠P = 180

∴∠MON+2∠P= 180。